Đáp án:
Gọi K là trực tâm tam giác ABC
$\begin{array}{l}
a)A\left( { - 1;3} \right);B\left( {3;5} \right);C\left( {2;2} \right)\\
\Rightarrow \overrightarrow {BC} = \left( { - 1; - 3} \right);\overrightarrow {AC} = \left( {3; - 1} \right)\\
K\left( {x;y} \right)\\
\Rightarrow \overrightarrow {AK} = \left( {x + 1;y - 3} \right)\\
\overrightarrow {BK} = \left( {x - 3;y - 5} \right)\\
Do:AK \bot BC;BK \bot AC\\
\Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AK} .\overrightarrow {BC} = 0\\
\overrightarrow {BK} .\overrightarrow {AC} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1.\left( {x + 1} \right) + \left( { - 3} \right).\left( {y - 3} \right) = 0\\
3.\left( {x - 3} \right) - 1.\left( {y - 5} \right) = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- x - 1 - 3y + 9 = 0\\
3x - 9 - y + 5 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + 3y = 8\\
3x - y = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 2
\end{array} \right.\\
\Rightarrow K\left( {2;2} \right)
\end{array}$
$\begin{array}{l}
\Rightarrow K\left( {2;2} \right) \equiv C\\
\Rightarrow \Delta ABC \bot C\\
\Rightarrow H\left( {2;2} \right) \equiv C
\end{array}$
Vậy H(2;2)
b)
$AC = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {2 - 3} \right)}^2}} = \sqrt {10} $
